Q.Choose the correct statement: (A) Photons of light show diffraction whereas electrons do not show diffraction. (B) Electrons have momentum whereas photons do not have momentum. (C) Photons of light and electrons both exhibit dual nature. (D) All electromagnetic radiations do not have photons.

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.When two electrons A and B are accelerated through potential difference $V_A$ and $V_B$ respectively, their deBrogli wave lengths are in the ratio of 2:3, then the ratio of potential differences $\dfrac{V_A}{V_B}$ is (A) 3:2 (B) 2:3 (C) 4:9 (D) 9:4

   ›Reveal solutionSolution

   Since $\lambda\propto1/\sqrt V$ for an accelerated electron, a wavelength ratio of 2:3 inverts and squares to give a potential-difference ratio of 9:4.

   Concept and Intuition

   An electron accelerated through a potential difference $V$ gains kinetic energy $eV = \dfrac{p^2}{2m}$, so its momentum is $p=\sqrt{2meV}$, and its deBroglie wavelength is $\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}$. Wavelength is thus inversely proportional to the square root of the accelerating voltage.

   Step-by-Step Solution

   1. $\lambda \propto \dfrac{1}{\sqrt V}$ (same charge and mass for both electrons).
   2. $\dfrac{\lambda_A}{\lambda_B} = \sqrt{\dfrac{V_B}{V_A}} = \dfrac{2}{3}$ (given).
   3. Squaring: $\dfrac{V_B}{V_A} = \dfrac{4}{9}$. …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the potential difference used to accelerate electrons at rest is increased from 150 V to 600 V, then the percentage decrease in the de-Broglie wavelength associated with the electron is (A) 25 (B) 75 (C) 50 (D) 40

   ›Reveal solutionSolution

   The de Broglie wavelength scales as $1/\sqrt V$; quadrupling the accelerating voltage halves the wavelength, i.e. a 50% decrease.

   Concept and Intuition

   An electron accelerated from rest through a potential difference $V$ gains kinetic energy $eV=\dfrac{p^2}{2m}$, so its momentum is $p=\sqrt{2meV}$, and its de Broglie wavelength is $\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}\propto\dfrac{1}{\sqrt V}$.

   Step-by-Step Solution

   1. $\dfrac{\lambda_2}{\lambda_1}=\sqrt{\dfrac{V_1}{V_2}}=\sqrt{\dfrac{150}{600}}=\sqrt{\dfrac14}=\dfrac12$.
   2. So the new wavelength is half the original: $\lambda_2=0.5\,\lambda_1$. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Two identical parallel metal plates A and B, having fine holes at their centers are connected to a power supply as shown in the figure. An electron having energy 200 eV is directed to pass through these holes from A to B. The de Broglie wavelength of the electron when it comes out of plate B is [FIGURE] (two parallel metal plates A and B, each with a small central hole, connected across a 100 V power supply; an electron travels horizontally through the aligned holes from A to B) (A) 0.713 Å (B) 2.012 Å (C) 1.754 Å (D) 1.227 Å

   ›Reveal solutionSolution

   The electron's kinetic energy changes by exactly the 100 eV set by the plate-to-plate potential difference; working out its final energy (100 eV) and applying $\lambda = h/p$ gives 1.227 Å.

   Concept and Intuition

   An electron's de Broglie wavelength is $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mE}}$, where $E$ is its kinetic energy. When an electron of charge $-e$ moves through a region with potential difference $V$, its kinetic energy changes by $eV$ (gaining energy if accelerated, losing energy if decelerated by the field). Here, an electron already carrying 200 eV of kinetic energy passes between two plates connected across a 100 V supply; crossing this potential difference changes its energy by exactly $100\ \text{eV} = eV$.

   Step-by-Step Solution

   1. Initial kinetic energy of the electron: $E_1 = 200$ eV.
   2. The field between A and B (100 V across the plates) does work $eV = 100$ eV on the electron as it crosses from A to B; here it acts to decelerate the electron, so the final kinetic energy is $E_2 = E_1 - eV = 200 - 100 = 100$ eV.
   3. Use the standard relation for an electron's de Broglie wavelength in terms of its kinetic energy expressed as an equivalent accelerating voltage $V_{eq}$ (numerically equal to the energy in eV): $\lambda(\text{Å}) = \dfrac{12.27}{\sqrt{V_{eq}}}$. …

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4. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Two particles of masses 'm' and '2m' are falling from same height. Then the ratio of the de Broglie wavelengths on reaching ground is (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1

   ›Reveal solutionSolution

   Free-fall speed is mass-independent, so momentum (and hence de Broglie wavelength) scales directly with mass, giving a $2:1$ wavelength ratio.

   Concept and Intuition

   A classic point of free-fall kinematics: acceleration due to gravity is the same for all masses, so both particles hit the ground with identical speed $v=\sqrt{2gh}$. But momentum $p=mv$ does depend on mass, so the heavier particle has twice the momentum — and since de Broglie wavelength $\lambda=h/p$ is inversely proportional to momentum, the lighter particle has the longer wavelength.

   Step-by-Step Solution

   1. Both fall through height $h$: $v = \sqrt{2gh}$ for each, independent of mass.
   2. Momentum of mass $m$: $p_1 = mv$. Momentum of mass $2m$: $p_2 = 2mv$.
   3. $\lambda_1 = h/p_1$, $\lambda_2 = h/p_2$. …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.The kinetic energy of a sub atomic particle is increased by 8 times. The de Broglie wavelength of it becomes $x$ times the initial wavelength. The value of $x$ is (A) $\dfrac{1}{8}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2\sqrt{2}}$ (D) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   De Broglie wavelength varies as $1/\sqrt{KE}$; an 8-fold increase in kinetic energy shrinks wavelength to $1/(2\sqrt2)$ of its original value.

   Concept and Intuition

   The de Broglie wavelength is $\lambda = h/p$, and kinetic energy relates to momentum via $KE = p^2/(2m)$, so $p=\sqrt{2m\,KE}$. This means $\lambda$ is inversely proportional to the square root of kinetic energy — doubling KE doesn't halve the wavelength, it divides it by $\sqrt2$.

   Step-by-Step Solution

   1. $\lambda = \dfrac{h}{\sqrt{2m\,KE}}$.
   2. If $KE' = 8\,KE$: $\lambda' = \dfrac{h}{\sqrt{2m(8KE)}} = \dfrac{h}{\sqrt{8}\sqrt{2m\,KE}} = \dfrac{\lambda}{\sqrt8}$. …

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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The de Broglie wavelength of an electron in the third Bohr orbit of H-atom is (A) $3\pi \times 5.29$ pm (B) $4\pi \times 52.9$ pm (C) $6\pi \times 52.9$ pm (D) $2\pi \times 5.29$ pm

   ›Reveal solutionSolution

   This tests Bohr's quantization condition connecting orbit circumference to the de Broglie wavelength. For the third orbit of hydrogen, $\lambda=6\pi\times52.9$ pm.

   Concept and Intuition

   Bohr's postulate that angular momentum is quantized, $mvr=\dfrac{nh}{2\pi}$, is equivalent (via the de Broglie relation $\lambda=h/mv$) to saying that exactly $n$ de Broglie wavelengths fit around the orbit's circumference: $2\pi r_n=n\lambda$. So once the orbit radius $r_n$ is known, the electron's wavelength in that orbit follows directly from the circumference divided into $n$ equal parts.

   Step-by-Step Solution

   1. Bohr quantization $\Rightarrow$ standing-wave condition: $2\pi r_n=n\lambda\ \Rightarrow\ \lambda=\dfrac{2\pi r_n}{n}$.
   2. Bohr radius of H-atom: $r_n=52.9\,n^2$ pm (since $Z=1$).
   3. Substitute:

   $\lambda=\frac{2\pi(52.9\,n^2)}{n}=2\pi(52.9)\,n$

   4. For the third orbit, $n=3$:

   $\lambda=2\pi(52.9)(3)=6\pi\times52.9\ \text{pm}$

   Common Mistakes …

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7. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.If A, B and C represent Planck's constant, mass and velocity respectively, then the dimensional formula of $\dfrac{A}{BC}$ is (A) $[M^0 L^0 T^1]$ (B) $[M^1 L^0 T^0]$ (C) $[M^0 L^1 T^0]$ (D) $[M^1 L^1 T^{-1}]$

   ›Reveal solutionSolution

   This tests dimensional analysis of Planck's constant combined with mass and velocity. The answer is (C) $[M^0 L^1 T^0]$.

   Concept and Intuition

   Dimensional formulas let us combine physical quantities symbolically without worrying about units, by tracking powers of mass $[M]$, length $[L]$, and time $[T]$. Planck's constant appears in $E = h\nu$, so its dimension is energy divided by frequency. Since energy has dimension $[ML^2T^{-2}]$ and frequency has dimension $[T^{-1}]$, Planck's constant $h$ has dimension $[ML^2T^{-1}]$. Dividing this by mass times velocity systematically cancels out mass and one power of length-per-time, leaving a pure length dimension.

   Step-by-Step Solution

   1. Dimension of Planck's constant $A = h$: from $E=h\nu$, $[h] = \dfrac{[E]}{[\nu]} = \dfrac{ML^2T^{-2}}{T^{-1}} = [ML^2T^{-1}]$.
   2. Dimension of mass $B = [M]$.
   3. Dimension of velocity $C = [LT^{-1}]$. …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.An electron of specific charge $\dfrac{e}{m}$ enters an electric field, $\overline{E} = -E_0\hat{i}$ at a time t = 0 with an initial velocity $v\hat{i}$. If $\lambda_0$ is its initial de Broglie wavelength, then its de Broglie wavelength at a time 't' is (A) $\dfrac{\lambda_0}{\left(1+\dfrac{eE_0t}{mv}\right)}$ (B) $\lambda_0\left[1+\dfrac{eE_0t}{mv}\right]$ (C) $\dfrac{\lambda_0}{\left(1-\dfrac{eE_0t}{mv}\right)}$ (D) $\dfrac{\lambda_0E_0}{mv}$

   ›Reveal solutionSolution

   The electric field accelerates the electron in the same direction it is already moving, so its speed and hence its de Broglie wavelength change as $\lambda=\lambda_0/(1+eE_0t/mv)$.

   Concept and Intuition

   The de Broglie wavelength is $\lambda = h/(mv)$ — inversely proportional to speed. To find $\lambda(t)$ we need $v(t)$, which requires knowing whether the field speeds up or slows down the electron. The electron carries charge $-e$ (with $e>0$ the elementary charge), and the field is $\vec E=-E_0\hat i$. The force is $\vec F=q\vec E=(-e)(-E_0\hat i)=+eE_0\hat i$ — pointing along $+\hat i$, the same direction as the initial velocity $v\hat i$. So the electron accelerates (speeds up), and its wavelength must shrink.

   Step-by-Step Solution

   1. Initial wavelength: $\lambda_0 = \dfrac{h}{mv}$.
   2. Force on electron: $\vec F = (-e)\vec E = (-e)(-E_0\hat i)=eE_0\hat i$, so acceleration $a=\dfrac{eE_0}{m}$ along $+\hat i$.
   3. Velocity at time $t$ (uniform acceleration, initial velocity $v$ along $+\hat i$):

   $v(t) = v + \frac{eE_0}{m}t = v\left(1+\frac{eE_0t}{mv}\right)$

   4. New wavelength: $\lambda(t) = \frac{h}{m\,v(t)} = \frac{h}{mv\left(1+\frac{eE_0t}{mv}\right)} = \frac{\lambda_0}{1+\frac{eE_0t}{mv}}$ …

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9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.If the kinetic energy of a particle having wavelength $x$ Å is increased to three times, its de Broglie wavelength (in Å) is : (A) $3x$ (B) $\sqrt{3}x$ (C) $\dfrac{x}{\sqrt{3}}$ (D) $\dfrac{x}{3}$

   ›Reveal solutionSolution

   Since $\lambda \propto 1/\sqrt{KE}$, tripling the kinetic energy shrinks the de Broglie wavelength by a factor $\sqrt3$.

   Concept and Intuition

   The de Broglie wavelength of a particle is $\lambda = h/p$, and kinetic energy relates to momentum via $KE = p^2/2m$, so $p=\sqrt{2m\cdot KE}$. Hence

   $\lambda = \frac{h}{\sqrt{2m\,KE}} \propto \frac{1}{\sqrt{KE}}$

   Larger kinetic energy means larger momentum, and larger momentum always means a shorter wavelength.

   Step-by-Step Solution

   1. Initially $\lambda_1 = x$ at kinetic energy $KE_1$.
   2. New kinetic energy $KE_2 = 3\,KE_1$.
   3. $\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{KE_1}{KE_2}} = \sqrt{\dfrac{1}{3}} = \dfrac{1}{\sqrt3}$.
   4. So $\lambda_2 = \dfrac{x}{\sqrt3}$.

   Common Mistakes …

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10. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127°\mathrm{C}$ and $352°\mathrm{C}$ is (A) $5:3$ (B) $3:2$ (C) $3:4$ (D) $5:4$

    ›Reveal solutionSolution

    The de Broglie wavelength of a thermal particle scales as $1/\sqrt{T}$; converting to Kelvin and taking the ratio gives $5:4$.

    Concept and Intuition

    A 'thermal' neutron's kinetic energy comes from thermal agitation, so on average $E\propto k_BT$. Since the de Broglie wavelength is $\lambda=h/p=h/\sqrt{2mE}$, and $E\propto T$, we get $\lambda\propto1/\sqrt{T}$ — hotter neutrons move faster and so have shorter wavelengths.

    Step-by-Step Solution

    1. Convert both temperatures to Kelvin: $T_1=127+273=400$ K, $T_2=352+273=625$ K.
    2. Since $\lambda\propto1/\sqrt{T}$: $\dfrac{\lambda_1}{\lambda_2}=\sqrt{\dfrac{T_2}{T_1}}=\sqrt{\dfrac{625}{400}}$. …

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11. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.The de Broglie wavelength associated with an electron accelerated through a potential difference of $\dfrac{200}{3}$ V is nearly (A) $25\ \overset{\circ}{A}$ (B) $2.5\ \overset{\circ}{A}$ (C) $15\ \overset{\circ}{A}$ (D) $1.5\ \overset{\circ}{A}$

    ›Reveal solutionSolution

    This tests the de Broglie wavelength formula for an electron accelerated through a potential difference; the answer is about 1.5 Å.

    Concept and Intuition

    An electron accelerated through potential $V$ gains kinetic energy $eV = \dfrac{p^2}{2m}$, giving momentum $p = \sqrt{2meV}$. Its de Broglie wavelength is $\lambda = h/p$. Substituting the constants gives the handy numerical formula $\lambda(\overset{\circ}{A}) = \dfrac{12.27}{\sqrt{V}}$ when $V$ is in volts — this is the standard shortcut used for electron-diffraction-type problems.

    Step-by-Step Solution

    1. Accelerating potential: $V = \dfrac{200}{3} \approx 66.67\,\mathrm{V}$.
    2. $\sqrt{V} = \sqrt{66.67} \approx 8.165$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.An electron of mass 'm' with initial velocity $\bar{v} = v_0\hat{i}$ ($v_0 > 0$) enters in an electric field $\bar{E} = -E_0\hat{i}$ [$E_0$ is constant > 0] at t = 0. If λ is its de-Brogli wavelength initially, then the de-Brogli wavelength after time 't' is (A) $\dfrac{\lambda}{1+\dfrac{eE_0t}{mv_0}}$ (B) $\dfrac{\lambda}{\left(1-\dfrac{eE_0t}{mv_0}\right)^2}$ (C) $\left(1+\dfrac{eE_0t}{mv_0}\right)\lambda$ (D) $\left(1+\dfrac{eE_0t}{mv_0}\right)^2\lambda$

    ›Reveal solutionSolution

    The field accelerates the electron in the direction it's already moving, increasing its speed

    (and hence decreasing its de Broglie wavelength) by a linear factor in time.

    Concept and Intuition

    The electron carries charge $-e$. In a field $\bar E = -E_0\hat i$, the force is $\bar F = q\bar E = (-e)(-E_0\hat i) = +eE_0\hat i$ — pointing in the $+\hat i$ direction, which is the same

    direction as the electron's initial velocity $v_0\hat i$. So the electron is accelerated forward

    (sped up), not decelerated.

    Step-by-Step Solution

    1. Acceleration: $a = F/m = eE_0/m$, directed along $+\hat i$ (same as $v_0$).
    2. Velocity at time t: $v(t) = v_0 + at = v_0 + \dfrac{eE_0}{m}t = v_0\left(1+\dfrac{eE_0t} {mv_0}\right)$.
    3. De Broglie wavelength $\lambda = h/(mv)$, so $\lambda \propto 1/v$.
    4. Initially $\lambda = h/(mv_0)$. At time t: $\lambda(t) = \dfrac{h}{mv(t)} = \dfrac{h/(mv_0)}{1+\dfrac{eE_0t}{mv_0}} = \dfrac{\lambda}{1+\dfrac{eE_0t}{mv_0}}$.

    Common Mistakes

    • Getting the sign of the force wrong (thinking the electron's negative charge in a $-\hat i$ …

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